I feel like the solution to this is painfully obviously, but I just can't seem to get it right... Any help would be greatly appreciated.

$\displaystyle f(x)=\int_{3}^{x^3}{t^2}{dt}$

It wants to know $\displaystyle f'(x)$. Wouldn't that just be $\displaystyle x^2$ then? Since the derivative cancels out the integral? What is messing me up is the limits, since I am pretty sure they make it something other than $\displaystyle x^2$.

I feel like the solution to this is painfully obviously, but I just can't seem to get it right... Any help would be greatly appreciated.

$\displaystyle f(x)=\int_{3}^{x^3}{t^2}{dt}$

It wants to know $\displaystyle f'(x)$. Wouldn't that just be $\displaystyle x^2$ then? Since the derivative cancels out the integral? What is messing me up is the limits, since I am pretty sure they make it something other than $\displaystyle x^2$.

What is true is that the derivative of $\displaystyle \int_a^x f(t) dt$ is f(x). If you have $\displaystyle F(x)= \int_a^{g(x)} f(t)dt$, for some function g(x), make a "change of variable". Let u= g(x) so that this becomes $\displaystyle F(u)= \int_a^u f(t)dt$. Now we have $\displaystyle \frac{dF}{dx}= \frac{dF}{du}\frac{du}{dx}$. Since u= g(x), $\displaystyle \frac{du}{dx}= \frac{dg}{dx}$ and so $\displaystyle \frac{dF}{dx}= f(g(x))\frac{dg}{dx}$

Here, "f(t)" is $\displaystyle t^2$ and g(x) is $\displaystyle x^3$ so the derivative is $\displaystyle (x^3)^2(3x^2)= 3x^8$.